solid mechanics study guide
TRANSCRIPT
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Solid Mechanics Study Guide Introduction Fundamental Types of Equations
1. equilibrium 2. geometry of deformation 3. material behavior
Geometry of Deformation in Problems:
definitions of extensional strain and shear strain simplifications and idealizations
(“rigid” member, “fixed” support, displacements are small, planes sections remain plane) connectivity of members or geometric compatibility boundary conditions or other constraints
Stress Definitions stress – the distribution of a force over the area on which it acts and is expressed as force intensity homogeneous – material is the same throughout the bar crosssection – section perpendicular to the longitudinal axis of the bar prismatic – straight and cross‐section does not change along the axis of bar uniaxial bar – a bar with only one axis Units English: lb/in2 = psi
kip/in2 = ksi
SI: N/m2 = Pa (Pascal) kN/m2 = kPa 106 N/m2 = MPa 109 N/m2 = GPa
Normal Stress – stress acting perpendicular to the cross‐section
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σ = FA
assume force is evenly distributed over cross‐section, except near the point of application deformation is uniform throughout loads must act through the centroid of the cross‐section σ > 0: tensile σ < 0: compressive
axial stress – normal stress that is uniform over a cross‐section bearing stress – special case of compressive normal stress (when one object bears on another)
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σ b = PAb
= Ptd
A = projected area where bearing pressure is applied P = bearing force
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Shearing Stress – stress acting tangential to the surface of a material element
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τ = FA
commonly found in bolts, pins, and rivets do NOT assume τ is uniform over the cross‐section max value of τ may be considerably greater than the average
direct shear – caused by forces that act parallel to a surface of some part (ex sheet‐metal punch)
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τ = F(πd)t
Single Shear:
Double Shear:
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τ =F 2A
= F2A
Equilibrium:
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τ1 = τ 2
a shear stress must be balanced by three other stresses shear stresses on opposite (parallel) faces are equal in magnitude
and opposite in direction shear stresses on adjacent (perpendicular) faces are equal in
magnitude and both point toward or away from each other Stress on an Oblique Plane
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σ = PA0cos2θ
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τ = PA0sinθ cosθ
force can be resolved into components:
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F = Pcosθ and
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V = Psinθ area of the section is
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A0 = Aθ cosθ normal force applied to a bar on an inclined section produces a combination of shear and
normal stresses max σ at 0° and max τ at 45°
Strain Normal (Extensional) Strain – deformation per unit length of a member under axial loading
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ε = ΔLL
3
dimensionless quantity but can be expressed as in/in, μin/in (read as microinch per inch), mm/mm, or μ (read as microstrain)
magnitude of extensional strain is usually quite small δ = small displacement
Thermal Strain – changes in temperature produces expansion or compression causing strain
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εT =αΔT
α = coefficient of thermal expansion constrained (unable to deform freely) = thermal stress free to deform = thermal strain ‐‐> ε = εT εT > 0: expansion εT < 0: contraction
Shear Strain – change in angle between two originally perpendicular line segments
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γ = π2 −θ
θ = inner angle after deformation result of shear stress τ dimensionless but usually expressed in the same units as extensional strain or in radians
Mechanical Properties of Materials Hooke’s Laws Young’s Modulus (modulus of elasticity E) – ratio of stress to strain
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σ = Eε
Shear Modulus (modulus of rigidity G) – ratio of shear stress to shear strain
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τ =Gγ NOTE: typical units for E and G are ksi or GPa Design Properties Strength:
yield strength – highest stress the material can withstand without undergoing yielding ultimate strength – maximum value of stress the material can withstand fracture stress – value of the stress at fracture
Stiffness: ratio of stress to strain interest primarily in linearly elastic region and therefore represented by value E
Ductility: ductile material – undergoes large permanent strains before failure brittle material – fails with little elongation after yield stress
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Stress‐Strain Diagram stress and strain are material properties load and dimension are specimen properties stress‐strain diagrams of various materials vary widely
Linear:
no permanent strain (safe zone) slope = E E has units of force per unit area
Yielding:
begins at yield stress σY slope rapidly until it is near horizontal strain is permanent (never go back to zero strain at this point) large strain increase with little to no stress increase
Strain‐Hardening:
stress begins to increase until it reaches the point of ultimate stress σu after undergoing large deformations it changes its crystalline structure material has increased resistance to applied stress (appears harder)
Necking: specimen begins to neck‐down loading beyond σu results in decreased load supported an eventually rupture stress appears to drop because graph does not take decreased cross‐section into account
Offset Method
used for some materials where there is not a clear yield stress draw a line with slope E and offset by a specified amount of strain σY is given by the intersection of this line with the stress‐strain curve commonly offset by a value of 0.002 (or 0.2%)
Elastic vs. Plastic elastic – strain caused by the application of load disappears when the load is removed, and it
returns to its original state (on stress‐strain diagram unloading path traces loading path) elastic limit – largest value of stress for which the material behaves elastically (basically σY) plastic – once yield stress obtained, when the load is removed, the stress ad strain decrease linearly
but do not return to their original state
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Axial Deformation Conditions:
axis of the member remains straight cross‐sections remain plane and perpendicular to the axis after deformation cross‐sections do not rotate about the axis
Uniform Axial Deformation
1. constant cross‐sectional area A 2. constant modulus of elasticity E 3. axial forces applied only at the ends
Elongation:
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e = FLAE
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F = ke
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k = AEL
Total Elongation:
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e = FLAE
+αLΔT
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F = k(e−αLΔT )
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e =δ = ΔL stiffness coefficient (k)– force required to produce a unit elongation (spring constant) Nonuniform Axial Deformation
Discrete Changes: total change in length is the summation of the change in
the length of each portion each time the internal force, area, or material changes you
need a new FBD
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e =FiLiAiEii
∑
Example:
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e =F1L1A1E
+F2L2A2E
+F3L3A3E
Continuous Changes:
continuously changing area or continuously changing force (such as a rod hanging under its under weight)
approximation because of the assumption that the stress distribution is constant over the cross‐section
good approximation for small variations
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e = F(x)A(x)E0
L∫ dx
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Types of Displacement Problems statically determinate – reactions and internal forces can be determined uniquely from FBD and
equations of equilibrium statically indeterminate – there are more unknown reactions than equations of equilibrium
Determining Equations Needed: M = R – N
N = number of equilibrium equations R = number of known reactions (or internal stresses) M = number of compatibility equations needed Example:
N = 1 (R1 – P – R2 = 0) R = 2 (R1, R2) M = 2 – 1 = 1
Solution Procedure: [Set up Fundamental Equations]
1. using FBD write down independent equilibrium equations 2. write an force‐deformation equation for each element:
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e = FL AE 3. use geometry of deformation to write down the compatibility equation in terms of the
element elongations e [Solve for Unknown Forces]
4. substitute force‐deformation equations into geometric compatibility equations 5. solve simultaneously the equilibrium equations and compatibility equations written in
terms of element forces [Solve for Displacements]
6. substitute the element forces into the force‐deformation equations 7. use geometry of deformation to relate element elongation to system displacements 8. review the solution
Design Considerations Factor of Safety:
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FS = ultimate loadallowable load
ultimate load – maximum load which a material can hold before failing allowable load – maximum load that the member is expected to see during its service lifetime
ultimate normal stress:
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σ u =PuAσ
ultimate shear stress:
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τ u =PuAτ
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Torsion Torsional‐Deformation Assumptions
axis remains straight and inextensible every cross‐section remains plane and perpendicular to the axis radial lines remain straight and radial as the cross‐section rotates about the axis
Variables: ϕ = angle of twist (in radians) τ = shear stress γ = shear strain ρ = radius (distance from axis) T = torque (moment) L = length of bar Ip = polar moment of cross‐section
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φ(x) =φ xL
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dφdx = twist rate
Shearing Strain
assume small ϕ and small γ for each element shear strain γ varies linearly with ρ the distance from the axis max shear strain occurs on the surface of the cylinder when ρ = r applies for all materials: elastic, linear, non‐linear, plastic, etc
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γ =ρφL
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= ρ dφdx
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γmax =r2φL
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γmin =r1φL
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γ =ρrγmax
Shearing Stress & Torque
use in simplest case – linearly elastic material recall Hooke’s Law for shear Ip = polar moment of inertia of the cross‐section of the bar
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τ =ρrGγmax
Polar Moment of Inertia:
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I p = ρ2A∫ dA for circular cross‐section ‐‐>
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I p = 12 π r
4 Torsion Formula:
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τ =TρIP
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Torque‐Twist Equation:
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φ =Ti LiGi IPii
∑
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φ =T LGIP
Inclined Planes
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σθ = τ sin2θ
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τθ = τ cos2θ
Maximums:
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τmax =σmax τmax occurs at θ = 0° and ±90° σmax occurs at θ = ±45° Power Transmission Shafts
shaft applies a torque T to another device and vise versa notice that the slower the speed the higher the torque for a given power work =
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Tφ
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P = Tω
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P = 2πf T Units: Power (P) Watt (W) = N‐m/s
horsepower (hp) = 550 ft‐lb/s
Torque (T) ft‐lb J = N‐m
Angular Speed (ω) rad/s
Frequency (f) Hz = rev/s
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Equilibrium of Beams
Types of Supports roller – prevents displacement in the transverse direction, but permits displacement in axial
direction and z‐rotation pin – prevents displacement in the axial and transverse directions, but permits z‐rotation cantilever (fixed end) – prevents displacement in the axial ad transverse directions, and also
prevents z‐rotation Relations Between F, V, and M
w = load V = shear M = bending moment
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dVdx
= −w
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dMdx
=V
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V = − w dx0
x∫ +V0
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M = V dx0
x∫ +M0
Pure Bending plane of bending – longitudinal plane of symmetry—which is loaded and supported symmetrically neutral surface (NS) – longitudinal plane perpendicular to the pane of bending that remains free of
strain as the beam deforms neutral axis (NA) – intersection of the neutral surface with the cross‐section deflection curve – axis (intersection of neutral surface and plane of bending) of the deformed beam cross‐sections – plane and perpendicular to the axis of the unformed beam—remain plane and
perpendicular to the deflection curve of the deformed beam
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Axial Strain since angles do not change there is no shear stress (τ = 0) origin set at neutral axis ρ = radius of curvature c = max distance between NA and upper or lower surface max compressive strain occurs on the upper surface max tensile strain occurs on the lower surface
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εx = −yρ
Above NA: Below NA:
+y contracts compression –εx
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εmax = − cρ
–y expands tension +εx
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εmax = cρ
Transverse Strain υ = Poisson’s ratio (material dependent) anticlastic curvature – the curvature in a direction perpendicular to the direction of bending
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εy = εz = −υεx radius of anticlastic curvature =
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ρυ
Normal (Flexural) Stress
assume the beam is made of a linear‐elastic material normal stress varies linearly with the distance from the NS
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σ x = Eεx = −Eyρ
NOTE: the x‐axis passes through the centroid of each cross‐section of the undeformed beam
Moment‐Curvature Equation:
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M = EIρ Flexure Formula:
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σ x = −MyI
Shear Stress and Shear Flow
longitudinal shear stress balances the normal force F1—the resultant of the flexural stress on cross‐sectional area a above y1
transverse shear stress at an arbitrary level y in the cross section can be calculated by determining the longitudinal shear stress
no horizontal shear stress at top and bottom
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Horizontal Shear Force:
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ΔH = VQIΔx
Shear Flow:
q ‐‐> shear force per unit length
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q = ΔHΔx
= VQI
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Q = ′ A ′ y
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Q = first moment of the area with respect to the NA
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′ A = cross‐sectional area above level y
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′ y = vertical distance from NA to the centroid of area A Shear‐Stress:
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τ = VQIt
τ = average transverse shear stress at level y I = moment of inertia of the entire cross‐section t = width of the cross‐section at level y (perpendicular to shear flow)
Beams with Flanges: vertical shear stresses are larger in the web than the flange usually only calculate value in the web ignore effects of small fillets at corners flanges have large horizontal shear stresses
Centroids and Moments of Inertia Review
Shape
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y
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Ix Area Rectangle
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12 h
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112 bh
3
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bh Right Triangle
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13 h
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136 bh
3
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12 bh
Circle 0
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14 π r
4
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π r 2
Semicircle
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43π r
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18 π r
4
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12 π r
2
Centroid:
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y = ∫ y dAA
Moment of Inertia:
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Ix = ∫ y 2dA Design of Beams in Bending Common Shapes:
wide‐flange (W) I‐beam (S) channel (C)
Notation English: symbol nominal depth (in) ✕ weight per foot (lb) SI: symbol nominal depth (mm) ✕ mass (kg/m) Angles: L longest leg length ✕ leg length ✕ thickness
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NOTE: wood beam actual net dimensions (finish dimensions) are smaller than the nominal dimensions and should be used in structural calculations Safe Design elastic section modulus (S) – property of the cross‐sectional dimensions
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S = Ic
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σmax =M
maxS
Requirements: max stress no more than allowable stress (
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σmax ≤σ all )
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Smin =M
maxσ all
Procedure:
1. determine the value of σall 2. draw shear and moment diagrams 3. from diagrams, determine the max absolute bending moment 4. determine min allowable value of S 5. use Smin to determine best cross section dimensions (see Appendix for values)
Beam Deflection deflection curve – (elastic curve) characterized by the function
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υ(x) that gives the transverse displacement of the points that lie along the axis of the beam
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1ρ
= dθds
= d 2υdx 2
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θ = tan−1 dvdx
≈
dvdx
Assumptions: 1. angle θ is small 2. beam is linearly elastic
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EI ′ ′ ′ ′ υ = −w = dVdx
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EI ′ ′ ′ υ =V = dMdx
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EI ′ ′ υ = M EI = flexural rigidity = constant
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υ(x) = deflection
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′ υ (x) = slope of deflection curve
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δB =υ(L) = deflection at end
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θB = ′ υ (L) = slope at end
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Conditions used for: evaluating constants of integration boundary conditions – set of known conditions specified for behavior at the boundaries
deflections and slopes at supports known moment and shear conditions
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υA = 0
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θA = 0
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VB = 0
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MB = 0
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υA = 0
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MA = 0
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υB = 0
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MB = 0 continuity conditions – set of known conditions specified for behavior where two intervals meet
displacement continuity slope continuity symmetry conditions
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υ(C)AC =υ(C)CB
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θ(C)AC =θ(C)CB NOTE: statically indeterminate beams ‐‐> take advantage of the boundary conditions to solve Method of Superposition
deflection and slope produced by multiple loads acting simultaneously can be found be superimposing the deflections produced acting separately
reference the Appendix for common formulas method can be applied to both statically determinate and indeterminate beams
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Transformation of Stress State of Stress
all the stress vectors associated with all planes that pass through a particular point written in terms of the 6 normal and shear stress components only one unique state of stress at a point but with different representations depending on
the orientation of the axes used plane stress – state of stress in a body if stresses along the z‐axis disappear
Stress‐Transformation Equations
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τ ′ x ′ y = −σ x −σ y
2
sin2θ + τ xy cos2θ
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σ ′ x =σ x +σ y
2
+
σ x −σ y
2
cos2θ + τ xy sin2θ
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σ ′ y =σ x +σ y
2
−
σ x −σ y
2
cos2θ −τ xy sin2θ
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σ ′ x +σ ′ y =σ x +σ y
Principal Stresses and Max Shear Stresses
principal stresses – max and min normal stresses orientation of the principal planes determined by the angles θp (90° from each other) NO shear stress on the principal planes planes of maximum shear stress lie at ±45° from principal planes normal stress at planes of max shear is σavg
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σ avg =σ x +σ y
2
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R=σ x −σ y
2
2
+ τ xy2
Principal Stress: Max Shear Stress:
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cos2θp = ±
σ x −σ y
2
R
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sin2θp = ±τ xyR
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sin2θs = ±
σ x −σ y
2
R
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cos2θs = ±τ xyR
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σ p =σ avg ± R
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τ p = 0
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τ s = ±R
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σ s =σ avg
Mohr’s Circle
center =
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(σ avg, 0) radius = R
angle θ on the body ‐‐> angle 2θ on Mohr’s circle τ is positive downward angles turned in the same direction on both the body
and on Mohr’s circle—counterclockwise original x‐axis and y‐axis are represented by line X‐Y principal stresses at intersections of circle and σ‐axis max shear stresses lie directly above and below center every point on the circle corresponds to the stresses
σ and τ on a particular face
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(σ ′ x , τ ′ x ′ y )
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Constructing and Using a Mohr’s Circle: draw axes plot center point O
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(σ avg, 0) plot given stresses point X
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(σ x, −τ xy ) and point Y
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(σ y, τ xy ) draw a line connecting X and Y through the center draw a circle with diameter X‐Y calculate principle stresses and plot on the σ‐axis 2θ is the angle CCW from original state X‐Y to the σ‐axis
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φ = 2θp ± angle of rotation
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σ ′ x = Rcosφ +σ avg
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σ ′ y = Rcosφ −σ avg
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τ ′ x ′ y = Rsinφ Poisson’s Ratio
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υ = −εyεx
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εx =σ xE−υσ y
E−υσ zE
Poisson’s ratio – the ratio of lateral strain to axial strain when an axial force is applied to a bar the bar not only elongates but also shortens in the
other directions material specific property and dimensionless limits of υ: 0 < υ < ½ material properties E and G can be related using υ
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G = E2(1+υ)
ThinWalled Pressure Vessels Assumptions
constant gage pressure (p = internal – external) thickness much less than radius internal radius = r point of calculation far away from the ends no shear stress in the cutting planes radial stress σr is usually ignored since it is so much smaller than in‐plane stresses
Cylindrical Pressure Vessels Axial (Longitudinal) Stress: Hoop (Circumferential) Stress: Shear Stress:
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σ a =pr2t
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σ h =prt
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τ absmax =pr4t
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τmax =pr2t
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Spherical Pressure Vessels Normal Stress: Shear Stress:
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σ s =pr2t
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τmax =pr4t
Stress Due to Combined Loads
Stress Resultant Symbol Formula
normal force F
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σ = FA
torsional moment T
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τ =TρIP
bending moment M
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σ = −MyI
transverse shear force V
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τ = VQIt
Example:
Equ # Equilibrium Equation Stress at H 1
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∑Fx = 0
2
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∑Fy = 0
3
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∑Fz = 0
4
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∑Mx = 0
5
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∑My = 0
6
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∑Mz = 0
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Buckling of Slender Columns Stability of Equilibrium stable equilibrium: after being slightly displaced the object moves back to equilibrium unstable equilibrium: once displaced the object tends to move farther away from equilibrium neutral equilibrium: no visible tendency for the object to return to equilibrium or farther from it
What happens as P increases?
1. point A moves downward and the beam shortens 2. at a certain value, Pcr, the beam “buckles” and suddenly changes shape 3. new shape may not support the full load
buckling – loss of stability of equilibrium critical load (
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Pcr) – value of the load at which the transition from stable equilibrium to unstable equilibrium occurs
Euler’s Formula
consider an ideal pin‐ended column solve for the deflection curve of the buckled beam and use boundary conditions to find
constants in the general solution buckling mode function
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υ(x) represents the shape of the deflected column infinite values but lowest value for P at which buckling occurs corresponds to n = 1 r =
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I /A = radius of gyration buckling load can be increased by increasing the value of the cross‐sectional moment of
inertia I by using thin‐walled tubular members critical (buckling) stress depends on the slenderness ratio (L/r)—below a certain ratio the
yield stress determines failure and above it buckling determines failure Buckling Load: Buckling Mode: Buckling Stress:
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Pcr = π 2EIL2
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υ(x) =C sin π xL
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σ cr = π 2E(L / r)2
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Other End Conditions extend Euler’s formula to columns with other end conditions replace length L with and equivalent or effective length, Le effective length – length of a pin‐ended column having the same buckling load as the actual
column (physically, the distance between points of zero moment when deflected) effectivelength factor (K) – dimensionless coefficient that depends on the end condition
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Pcr = π 2EILe2
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Le = KL